CHEMISTRY HIGH SCHOOL

ANSWER ASAP PLEASE Which of the following objects is the most dense?
Object A: mass = 50 g, volume = 5 cm³
Object B: mass = 50 g, volume = 10 cm³
Object C: mass = 10 g, volume = 50 cm³
Object D: mass = 5 g, volume = 50 cm³

Answers

Answer 1
Answer: Density is equal to mass divided by volume so the densest object will be the object that has the largest mass in the smallest area. In this case object A is the densest with a density of 10g/cm^3. I hope this helps. Let me know if anything is unclear.

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If a cell is placed in salt water, water leaves the cell by _____. A. Osmosis B. Diffusion C. Active transport D. Phagocytosis
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How does CFL lamps help in conservation of energy?

Answers

CFL means compact fluorescent lamp

It can save five times its purchase price in electricity cost because it contains toxic mercury that is equivalent in working: electrons that are bound to mercury atoms are excited to states where they will radiate ultraviolet light as they return to lower energy level, this ultraviolet light will be converted into visible light as it strikes to the coating of the bulb. Like LED that is high-efficiency house lighting.

A well-tested explanation that explains a lot of observations isA. a hypothesis.
B. an inference.
C. a theory.
D. a controlled experiment.

Answers

A: Because a hypothesis is a educated guess
A well-tested explanation that explains a lot of observations is a theory.

Consider the mixture of ethanol, c2h5oh, and o2. how many molecules of co2, h2o, c2h5oh, and o2 will be present if the reaction goes to completion?

Answers

The reaction between ethanol and oxygen must be a combustion reaction to give carbon dioxide and water as products.

The reaction between ethanol, C_2H_5OH and oxygen, O_2 is as follows:

C_2H_5OH+O_2\rightarrow CO_2+H_2O

In order to balance the reaction (the number of atoms of each element in the reaction are same on both the sides that is product and reactant),  O_2 is multiplied by 3 in the reactant side and CO_2 is multiplied by 2 and H_2O is multiplied by 3 on product side.

So, the balanced reaction is:

C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O

From the balanced reaction it is clear that 1 mole of C_2H_5OH reacts with 3 moles of O_2 to give 2 moles of CO_2 and 3 moles of H_2O.

Since, 1 mole = 6.022* 10^(23) molecules (Avogadro's number)

So,

Number of molecules of C_2H_5OH = 6.022* 10^(23) molecules.

Number of molecules of O_2 = 6.022* 10^(23)* 3 = 18.066* 10^(23) molecules.

Number of molecules of CO_2 = 6.022* 10^(23)* 2 = 12.044* 10^(23) molecules.

Number of molecules of H_2O = 6.022* 10^(23)* 3 = 18.066* 10^(23) molecules.

Answer : The number of molecules of CO_2,H_2O,C_2H_5OH\text{ and }O_2 are 2, 3, 1 and 3 respectively.

Explanation :

The given balanced chemical reaction is,

C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O

Balanced chemical reaction is the reaction in which the number of atoms of an individual elements present on reactant side must be equal to the product side.

The species present on the left side of the right arrow is the reactant and the species present on the right side of the right arrow is the product.

From the balanced chemical reaction, we conclude that there are 1 molecule of C_2H_5OH, 3 molecules of O_2, 2 molecules of CO_2 and 3 molecules of H_2O.

Which is a product of a condensation reaction? 1. O2 2. CO2 3. H2 4. H2O

Answers

The correct answer is H2O, it is a product of a condensation reaction. It is a type of chemical reaction where two molecules combine to form a larger molecule together with a loss of a small molecule. Usually, this small molecule is water.

Answer:

H_(2)O is a product of a condensation reaction.

Explanation:

In a condensation reaction, two molecules are combined to produced a large molecules along with removal of a water molecule.

For an example, let's consider aldol condensation reaction.

In aldol condensation reaction, an enolate anion reacts and combines with an aldehyde or ketone molecule follwed by dehydration. In that dehydration step  , H_(2)O gets removed and an unsaturated carbonyl product is formed.

Usually, O_(2),CO_(2) and H_(2) are produced as a by product of a decomposition reaction.

Magnesium is an alkaline earth metal with an atomic number of 12. It’s in the third row in the periodic table. Which expression shows one way to find the number of valence electrons in one atom of magnesium?

Answers

Answer:

group # = # valence electrons

group 2= 2 valence electrons

Explanation:

Mg is in group #2 and subsequently has 2 valence electrons (electrons on the outermost shell)

After 32 days, 5 milligrams of an 80-milligram sample of a radioactive isotope remains unchanged. What is the half-life of this element?(1) 8 days (3) 16 days
(2) 2 days (4) 4 days

Answers

Unstable heavy atoms will undergo radioactive decay to produce stable species. The half life time of the isotope which undergone a decay of 75 mg in 32 days is 18 days.

What is half life time?

The half life time of a radioactive sample is the time taken to reduce it to half of the initial amount by decay.

The heavy unstable material have very short half life and they will easily undergoes radioactive decay by emitting certain radiation.

Radioactive decay is a firs order reaction and have the equation to find the radioactive constant as follows:

\lambda = (1)/(t) log([Ni])/([Nt])

Where, t is the time of decay and Ni and Nt be the initial and final amount respectively.

It is given that 5 mg is remaining out of 80 mg after 32 days. Thus the radioactive constant is calculated as follows:

\lambda = (1)/(32 days ) log (80)/(5) \n \n = 0.0376. days ^(-1)

Now the half life time of the decay is calculated as below:

t(1/2) = 0.693 /decay constant

      = 0.693/0.0376

      = 18 days

Therefore, the half life time of the isotope which undergone a decay of 75 mg in 32 days is 18 days.

To find more about radioactive decay, refer the link below:

brainly.com/question/1770619

#SPJ5
m=m_0 * ((1)/(2))^(t)/(t_(1/2))
m - the mass that remains unchanged, m₀ - the inital mass, t - the time of decay, t1⁄2 - the half-life

t=32 \ days \n m=5 \ mg \n m_0 = 80 \ mg \n \n 5 = 80 * ((1)/(2))^(32)/(t_(1/2)) \ \ \ \ \ \ \ |/ 80 \n (5)/(80)= ((1)/(2))^(32)/(t_(1/2)) \n (1)/(16)=((1)/(2))^(32)/(t_(1/2)) \n ((1)/(2))^4=((1)/(2))^(32)/(t_(1/2)) \n 4=(32)/(t_(1/2)) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |* t_(1/2) \n 4t_(1/2)=32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |/ 4 \n t_(1/2)=8

The half-life is (1) 8 days.
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